As with any stoichiometry problem, we start with the balanced equation.
#"Cu" + "2AgNO"_3 → "Cu"("NO"_3)_2 + 2"Ag"#
Mass of Cu
#"Moles of AgNO"_3 = 1.000 color(red)(cancel(color(black)("g AgNO"_3))) × ("1 mol AgNO"_3)/(169.87 color(red)(cancel(color(black)("g AgNO"_3)))) = "0.005 887 mol AgNO"_3#
#"Moles of Cu" = "0.005 887" color(red)(cancel(color(black)("mol AgNO"_3))) × (1 "mol Cu")/(2 color(red)(cancel(color(black)("mol AgNO"_3)))) = "0.002 943 mol Cu"#
#"Mass of Cu" = "0.002 943" color(red)(cancel(color(black)("mol Cu"))) × ("63.55 g Cu")/(1 color(red)(cancel(color(black)("mol Cu")))) = "0.1871 g Cu" = "187.1 mg Cu"#
Mass of 999 Ag
#"Moles of Ag" = "0.005 887" color(red)(cancel(color(black)("mol AgNO"_3))) × "2 mol Ag"/(2 color(red)(cancel(color(black)("mol AgNO"_3)))) = "0.005 887 mol Ag"#
#"Mass of Ag" = "0.005 887" color(red)(cancel(color(black)("mol Ag"))) × "107.9 g Ag "/(1 color(red)(cancel(color(black)("mol Ag")))) = "0.6352 g Ag" = "635.2 mg Ag" #
#"Mass of 999 Ag" = 635.2 color(red)(cancel(color(black)("mg Ag"))) × "1000 mg 999 Ag"/(999 color(red)(cancel(color(black)("mg Ag")))) = "635.8 mg 999 Ag"#
You will need to add 187.1 mg of #"Cu"#, and you will get 635.8 mg of 999 #"Ag"#.
